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The pressure variation in a static fluid depends only on depth, which is represented by the equation: $$P=P_0+\rho_{water} gz\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$$

When an object is submerged in a static fluid, the pressure at a point below the object is not affected by the presence of the object, i.e., $$P_{below}=P_0+\rho_{water} gz\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(2)$$, in which z is the depth to that point.

My question is: in equation (2), why is the density of the water not replaced by the density of the object, i.e., $$P_{below}=P_0+\rho_{object} gh+\rho_{water} gz_{water }\space \space\space\space\space\space\space\space\space\space(3) $$, in which $h$ is the height of the object and $z_{water}$ is the depth to the top of the object.

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  • $\begingroup$ I wonder if you are studing hydrostatics, since I realize great confusion for simple facts. Take a look in my answer as user82794 therein : Proof of Archimedes Principle. $\endgroup$
    – Frobenius
    Aug 7, 2017 at 7:18
  • $\begingroup$ Could you explain further your answer? In the other answer you proved the Archimedean property, and in your work the density inside the fluid at any point is pgh; But I am looking for the explanation why that formula holds for a point below a submerged object $\endgroup$ Aug 7, 2017 at 8:01
  • $\begingroup$ "...in my work the static pressure (and not the density) inside the fluid is $\,\rho gh\,$; But..." Your Great Confusion : you consider the object like a truck weighting 5 tones and the fluid beneath it like a road surface on which this weight is applied. I apologize but you are away from understanding the basic and elementary principles of Hydrostatics and far far away from understanding Fluid Mechanics. $\endgroup$
    – Frobenius
    Aug 7, 2017 at 11:23

2 Answers 2

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OK , i am assuming that you mean that the object you are talking about is in equilibrium.

in this case the density of the object must be equal to the density of the liquid , so your equation (3) reduces to equation (2).

The density must be same as the difference in pressure above and below the object must balance the objects weight.

$ (\rho g h) A = Mg $

and

$ Mg = g(\sigma)(Ah) $

where h is the height of the object

A is the area of cross section

$\sigma $ and $\rho$ are the densities of the object and water respectively.

Thus ,

$\sigma=\rho$

Also note that the pressure in a liquid is the same at the same horizontal level (given that the liquid is not accelerating sideways)

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  • $\begingroup$ But if the object's density is not constant but its average is equal to that of water, then your argument still holds but equation (3) wouldn't reduce to (2). Correct me if I am wrong. Thanks. $\endgroup$ Aug 7, 2017 at 20:50
  • $\begingroup$ In case of variable density too your equation (3) would reduce to (2) . You would just have to do some integration while calculating the pressure due to the object $\endgroup$ Aug 8, 2017 at 17:09
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enter image description here

If your assumptions were correct, that drop of water would accelerate away, thus not a static situation.

(Let's just assume that object is fixed to that point somehow; I just didn't draw that)

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  • $\begingroup$ Doesn't it mean that a static situation wouldn't be possible in this case? $\endgroup$ Aug 7, 2017 at 20:46
  • $\begingroup$ @Geophysics no; it means a static situation where $P_1 \ne P_2$ would not be possible. It does not deny the possibility of those with different pressures. $\endgroup$
    – Carl Lei
    Aug 8, 2017 at 0:46

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